Mark Scheme (Results) January Pearson Edexcel International Advanced Level. Mechanics 3 (WME03/01)

Transcription

1 Mark Scheme (Resuts) January 04 Pearson Edexce Internationa Advanced Leve Mechanics 3 (WME03/0)

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3 Genera Marking Guidance A candidates must receive the same treatment. Examiners must mark the first candidate in exacty the same way as they mark the ast. Mark schemes shoud be appied positivey. Candidates must be rewarded for what they have shown they can do rather than penaised for omissions. Examiners shoud mark according to the mark scheme not according to their perception of where the grade boundaries may ie. There is no ceiing on achievement. A marks on the mark scheme shoud be used appropriatey. A the marks on the mark scheme are designed to be awarded. Examiners shoud aways award fu marks if deserved, i.e. if the answer matches the mark scheme. Examiners shoud aso be prepared to award zero marks if the candidate s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes wi provide the principes by which marks wi be awarded and exempification may be imited. Crossed out work shoud be marked UNLESS the candidate has repaced with an aternative response.

4 EDEXCEL GCE MATHEMATICS Genera Instructions for Marking. The tota number of marks for the paper is 75.. The Edexce Mathematics mark schemes use the foowing types of marks: M marks: method marks are awarded for knowing a method and attempting to appy it, uness otherwise indicated. A marks: Accuracy marks can ony be awarded if the reevant method (M) marks have been earned. B marks are unconditiona accuracy marks (independent of M marks) Marks shoud not be subdivided. 3. Abbreviations These are some of the traditiona marking abbreviations that wi appear in the mark schemes. bod benefit of doubt ft foow through the symbo wi be used for correct ft cao correct answer ony cso - correct soution ony. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: specia case oe or equivaent (and appropriate) dep dependent indep independent dp decima paces sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark 4. A A marks are correct answer ony (cao.), uness shown, for exampe, as A ft to indicate that previous wrong working is to be foowed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifesty absurd answers shoud never be awarded A marks. 5. For misreading which does not ater the character of a question or materiay simpify it, deduct two from any A or B marks gained, in that part of the question affected. 6. If a candidate makes more than one attempt at any question: If a but one attempt is crossed out, mark the attempt which is NOT crossed out. If either a attempts are crossed out or none are crossed out, mark a the attempts and score the highest singe attempt. 7. Ignore wrong working or incorrect statements foowing a correct answer.

5 Genera Notes From Chief Examiner Usua rues for M marks: correct no. of terms; dim correct; a terms that need resoving (i.e. mutipied by cos or sin) are resoved. Omission or extra g in a resoution is accuracy error not method error. Omission of mass from a resoution is method error. Omission of a ength from a moments equation is a method error. Omission of units or incorrect units is not (usuay) counted as an accuracy error. DM indicates a dependent method mark i.e. one that can ony be awarded if a previous specified method mark has been awarded. Any numerica answer which comes from use of g = 9.8 shoud be given to or 3 SF. Use of g = 9.8 shoud be penaised once per (compete) question. N.B. Over-accuracy or under-accuracy of correct answers shoud ony be penaised ONCE per compete question. In a cases, if the candidate ceary abes their working under a particuar part of a question i.e. (a) or (b) or (c), then that working can ony score marks for that part of the question. Accept coumn vectors in a cases. Misreads if a misread does not ater the character of a question or materiay simpify it, deduct two from any A or B marks gained, bearing in mind that after a misread, the subsequent A marks affected are treated as A ft.

6 Question Number Scheme Marks. 3 v = 8x 4 v = 8x 4 3 dv v = x dx F = 0.5 6x = 3x x = 4 F = 6 M A A Mdep A 5 M Notes for attempting to differentiate the expression for v - chain rue must be used on hs. x A for correct A for 6 Award both ony if work fuy correct Mdep for using NL with m = 0.5 to obtain an expression for F in terms of x Acso for F = 6 Aternatives: for the first 3 marks 3 dv = 8 x 4 x dx dv dv = x v = 6x dx v dx 3 dv dx = 8x 4 x dt dt 3 3 dv = 8 x 4 x 8 x 4 = 6 x dt M Must be a compete method to obtain acce in terms of x Arhs Ahs MAA Award as above

7 Question Number Scheme mg x = mg + x 4 x + x = 8 0 ( x )( x ) 4 + = 0 x = or 4 3 distance = + = 4 4 Notes Marks MA;M A M A Mdep A A 9 M A M A for the difference of eastic energy terms, not nec in a compete energy equation. for a correct difference for a work energy equation, oss of EPE = work done against friction(not dep on previous mark) for a fuy correct equation Mdep for re-arranging to a three term quadratic, dependent on the second M mark, or use the difference of squares to get a inear equation A for a correct 3 term quadratic, terms in any order Mdep for soving the resuting quadratic, usua rues. Dependent on a second and third M marks A for x= 4 x= need not be shown Acao and cso distance = 3 4

8 Question Number 3 Scheme 9 mu mg mg = 8 a ag u = 4 ag ag m m = mga ( cosθ ) 4 0 θ = 8 nearest degree M A A Marks M A A Mdep A 8 M A A Notes for NL aong the radius at the bottom or top. Must have forces and an acceeration 9 mu for a fuy correct equation ie mg mg = oe Must be at the bottom 8 a ag for obtaining u = 4 ag M for an energy equation from the bottom or top to the point where the speed is (this may be 0 v here and for the A marks). Must have a difference of KE terms and a GPE term. Aft for correct difference of KE terms or correct PE term (from bottom) Foow through their u. A for a competey correct equation ag Mdep for substituting v = and soving for θ Dependent on both previous M marks 0 Acao θ = 8 must be nearest degree. If candidates do the energy equation first, give those 3 marks for an equation with u (speed at ag bottom) and. The fina M mark wi then be for substituting ag u = and soving for θ. 0 4 If the radius is a throughout, treat as mis-read. If sometimes a and sometimes a mark each equation on it own merit.

9 Question Number 4(a) (b) 0 Scheme x π x π e dx = e 0 π = ( e ) PRINTED ANSWER M A Marks Acso 3 x x x M A π xe dx = π xe e d π x x = π e + e 0 = π e ( e ) 4 3 e = π e π 4 4 ( e 3) x = = π ( e ) ( e ) Mdep Aft Acao M A (7) 0

10 Notes for Question 4 A note about π : (a) is a "show that" so π must be incuded throughout (uness a put in at the end of (a), with a convincing argument for doing so). No answer given in (b), so aow the first 5 marks (as earned) without π provided either no π s or both π s appear for the fina marks. If the fina fraction has the denominator π ony, the ast 3 marks wi be ost (a) M for using V = π y dx= π e x dx A for correct integration, correct imits must be shown Acso for V π ( e ) and attempting the integration. imits not needed for this mark = * Must be seen in this form (b) M for attempting the integration of π xe x dx by parts - imits not needed yet. Aow if intention to integrate π xy dx is shown. A for a correct resut with or w/o imits (check signs carefuy) Mdep for attempting the next integra, imits not needed A ft for substituting the correct imits in their integra Acao 3 for e π 4 4 oe ( π ) xy dx M for using x = π y dx with their integras, must be the correct way up. A ( ) ( e 3) ( ) for x = oe must be in terms of e. Must have ony terms in each of the numerator e and denominator and no fractions in either.

11 Question Number Scheme Marks 5(a) k πr kπr r k πr + kπr r ( ) ( 3) ( 5) 3r 3r + 3r x 8 B B 3r 3r + 3 r. +.3 = 5x 8 9r = x 4 PRINTED ANSWER M Aft A (5) (b) R= W ; F = P 9r P.rsin α = W( 4 sinα rcos α) 9 P= W( 8 cot α) F = µ R (9 4cot α ) = µ 8 PRINTED ANSWER B M A A A MdepAcso (7) Notes (a) B for a correct ratio of masses B for correct distances of the c of ms of the two components, hopefuy from O, but can be from another point M for a moments equation about O or their chosen point. Must have three terms and be dimensionay correct Aft for a correct equation, foow through their ratio of masses and distances, but not :3:4 (from mass/unit vo) 9r Acso for x = * 4 Specia case: Using voumes: max B0BMAA (b)b for the two shown equations M for a moments equation about the point of contact AA Award A if eqn fuy correct; AA0 if one error 9 A for re-arranging to obtain P W = cot α 8 Mdep for using F = µ R together with the expression for P and the first two equations to obtain an expression for µ Acso for µ = ( 9 4cot α) * must be this form 8

12 Question Number 6(a) (b) Scheme (6 a) + (8 a) = (0 a) by Pythag (converse), APB = 90 o PRINTED ANSWER T sinα + T cosα = mrω T cosα T sinα = mg r = 8asina 3 4 sin α = 5 or cosα = 5 3m soving, T = (3aω 5 g) 5 5g T 0 ω = 3a Marks M A () M A M A M A B M M A π 3a max time = = π ω 5g PRINTED ANSWER MA (3) 5

13 Notes for Question 6 (a) M for squaring the sides and showing they fit Pythagoras' theorem or ratio of sides 3:4:5 or use the cosine rue Acso for stating that (the converse of) Pythagoras' theorem shows that APB = 90 * or appropriate concusion for their method (b) M for NL horizontay. There must be two tensions, both resoved, but may be the same, and an acceeration (either form accepted here) Sine/cos interchange is an accuracy error. A for any two correct terms A for the third correct term. Acceeration must be in the form mrω and tensions must be different for both these marks to be awarded M for resoving verticay. Again, two tensions, both resoved but may be the same, and sine/cos interchange is an accuracy error. A for a fuy correct equation with different tensions. M for finding the radius as r = 8asin a or 8acosa A for r = 8asina May not be shown expicity B for a correct vaue for sin α or cosα Mdep for soving to obtain an expression for T in terms of mgaω,,,. Dependent on a M marks above and two different tensions. Or making T = 0 in the above equations and soving for ω Mdep for using T 0 in their expression for T to obtain an expression for ω in terms of g and a Dependent on the previous M mark T < 0 gets M0 5g A for ω min = oe 3a M for using π with their ω to obtain the maximum time ω Acso for max time = 3a π 5g

14 Question Number 7 (a) (b) Scheme 8mge = mg e= SHM, period p 8 mg T = mx 8mg mg ( x 8 ) = mx 8g x = x 8g PRINTED ANSWER Marks M A () M A Mdep A A Acso (6) (c) (d) a = - = u = (( ) - ( ) ) 8g u = g B M A A (4) x= -acosωt x = aωsinωt 9g 3 8g 8g = sin t 3 8 8g = sin t π t = 6 8g M A Mdep A (4) 6

15 Notes for Question 7 (a) M for Hooke's aw and equating tension to weight Acao for e= 8 (b) M for NL verticay, weight and tension needed, x or a for acceeration here A for a correct equation with x or a Mdep for using HL to repace the tension with an expression in terms of x Dependent on the previous M mark Must have x now A for this equation correct 8g A for re-arranging to get x = x oe Acso for the concusion SHM and the period π * 8g (c) 3 B for using the information in the question to obtain amp = 8 M for using v = ω ( a x ) with their ω and a A for a correct, unsimpified expression for u in terms of and g Acao for u = g By energy: B for EPE, M equation, A correct equation, A answer (d) M for using x = aωsinωt (or v instead of x ) with their a and ω and the given speed A for a fuy correct equation Mdep for soving their equation must use radians π Acao for t = or oe. (if sub for g seen, must be or 3 sf) 6 8g 8g Aternative for (d): v = ω a x with their ω and a and the given speed M Use ( ) 3 7 x = 3 or x = oe 6 56 A Use x= acosωt with their x, ω and a and sove in radians Mdep π t = or g 8g oe. (if sub for g seen, must be or 3 sf) Acao

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